A note on the proof of Bertrand's theorem

A NOTE ON THE PROOF OF

BERTRAND'S THEOREM

Vladimir Jovanovi¢

Abstrat. In this paper well a ommon gap inthe proof of Bertrand'

theorem present both the inBertrand's originalpaper Théorème relatif au

movementd'unpointattiréversunentrexeandintheArnold'sbook

Math-ematialmethodsof lassial mehanis, byprovidingmissing detailswhih

pertaintotheproblemofhowtosingleoutelastiandgravitationalpotentials

amongthepowerlawones.

1. Introdution

Motion of a partile with a unit mass in a entral eld is desribed by the

Newton's equations

(1.1)

¨

−

→

_r

₌

₋

∂U

∂

−

→

r

,

where

−

→

_r

₌

−

→

_r

_(t)

_∈

_R

3

istheradiusvetorofthepartileat time

t

,and

U

=

U

(r)

is the potential energy or simply the potential; here

r

=

|−

→

_r

_|

is the Eulidean

norm of the vetor

−

→

_r

. It is a ommon fat that everymotion in a entral eld

atually ours in a xed plane. If we denote by

π

the plane in whih a given motion ours, then we an write

−

→

_r

₌

_r(cos

_ϕ

−

→

_e

₁

_{+ sin}

_ϕ

−

→

_e

₂

₎

, where

−

→

e

1

,

−

→

e

2

is an

orthonormal basis of

π

, and

(r, ϕ)

are polar oordinates in this plane. Let us denote

−

→

e

r

=

−

→

r

r

= cos

ϕ

−

→

e

1

+ sin

ϕ

e

→

−

2

. Then (1.1)anberewrittenas (1.2)

¨

−

→

_r

₌

₋

_U

′

(r)

−

→

e

r

.

If

−

→

e

ϕ

=

−

sin

ϕ

−

→

e

1

+ cos

ϕ

−

→

e

2

, then

˙

−

→

_e

_r

_{= ˙}

_ϕ

−

_e

→

_ϕ

and

˙

−

→

e

ϕ

=

−

ϕ

˙

−

→

e

r

. Therefore,

−

→

_r

₌

_r

−

→

_e

_r

and(1.2)imply

(1.3)

r

¨

−

r

ϕ

˙

2

₌

−

U

′

(r),

2 ˙

r

ϕ

˙

+

r

ϕ

¨

= 0.

Fromtheseond equationin(1.3)follows

r

2

ϕ

˙

=

M,

2010MathematisSubjetClassiation: 70B05,34A34.

foraertainonstant

M

. Therstequationin (1.3)wetransforminto

(1.4)

r

¨

=

−

V

′

(r),

where

V

(r) =

U

(r) +

M

2

2r

2

is the amended potential. From (1.4) follows that the

energyofthesystem(1.1)isaonstantofthemotion:

1

2

r

˙

2

₊

_V

_{(r) =}

_E.

Ouronsiderationswillhavemuhtodowithirularorbits: thesearesolutionsof

(1.1) with

r

=

ρ

onstant. From(1.4)wededuethat

V

′

(ρ) = 0

isneessaryand suient ondition for

r

=

ρ

to bea irular orbit. Assume that

V

′′

(ρ)

>

0

. In [3, subsetions12.6and12.7℄is developedatheorywhihassertsthat foragiven

energy level

E > V

(ρ)

lose to

V

(ρ)

, the solutions of (1.4) periodially osillate in theinterval

[r

min

, r

max

]

∋

ρ

(see Figure 1). The position of the partile, when

r

=

r

min

isalledaperienter andfor

r

=

r

max

anapoenter. Theangle

Φ

between twoonseutiveperientersandapoenters,thesoalledapsidalangle,isgivenby

(see[2℄)

(1.5)

Φ =

Z

r

max

r

min

M/r

2

p

2(E

−

V

(r))

dr.

A simplebutfundamental observationisthatanorbitof (1.1)islosedifand

onlyif

Φ/π

∈

Q

.

r

r

min

r

max

E

V

Figure 1

Already Newton knew that all bounded orbits for elasti and gravitational

potentials are ellipses (see [9℄). However, if there are other entral potentials

whih give rise to losed orbitsunder arbitrary initial onditions(provided they

are bounded) hadbeenunknown for almost twoenturiesuntil Bertrand in 1873

published hispaper[4℄inwhihheprovedthat theanswerwasno. Thisassertion

istodayknownasBertrand's theorem.

A lotofeorthasbeenspenttoextendBertrand'stheorem tothegeometries

and spaesotherthantheEulideanthreedimensionalone: forexample,in [8℄it

isprovedfortheLobahevskyandspherialgeometry,andin[5℄,[10℄ forsurfaes

ofrevolution. LetusstressthatallthesegeneralizationsarebaseduponBertrand's

originalproof. Forfurtherdevelopmentin thisdiretionsee[11℄.

Some authors used quite dierent tools to handle the problem, in partiular

perturbation theory of integrableHamiltonian systems. For example,Albouy [1℄

and Féjoz,Kazmarek[6℄ appliedBirkhonormalforms alulatedalongirular

Bertrand's original proof anbedivided into three steps: rst, he alulated

thelimitoftheapsidalanglewhentheorbitapproahestheirularone,andthen

employingontinuousdependene oftheapsidalangleonorbitsheestablishedits

independene onenergy,whihonsequentlyimpliesthattheangleisequaltothe

aforementionedlimit. In the seond step is used the fat that the apsidal angle

is independent on momentum aswell, and, as a onsequene, one arrivesto the

powerlaw forthe potential:

U

(r) =

ar

α

. Thethird stepdealswith the problem

of howto singleoutthegravitationalandelasti potentialsamongthe powerlaw

ones. Bertrand resolvedthis stepessentiallyby takingsome speial limits in the

initial form ofthe apsidalangle. This stepofthe Bertrand'sproofis alsotreated

inArnold'sbook[2℄. However,inbothasesdetailslakanditturnsoutthatthey

arenotsoeasytoprove. Thisisexatlythepurposeofthepaper: toprovidesome

details missing in[4℄ and[2℄in theproofofthis step. Itis theontentofLemma

(2.1).

An originalapproahto thethird stepbasedonthefrationalalulusanbe

foundin [7℄.

2. Formulation and proofof Lemma

Lemma 2.1. If

U

(r) =

ar

α

and

a >

0

,then

lim

E

→∞

Φ =

π

2

,

for all

α >

0

.

Proof. Letusintrodueanewvariable

x

by

M

r

=

x.

Dene thefuntion

W

by

W

(x) =

V

(r)

,thatis

W

(x) =

1

2

x

2

₊

_U

M

x

=

1

2

x

2

₊

1

2

kx

−

α

_,

where

k

= 2aM

α

. Let

x

1

=

M/r

max

and

x

2

=

M/r

min

. Following[2℄weapplythe substitution

x

=

yx

2

in (1.5)andget

Φ =

Z

1

y

1

dy

p

H

(y)

,

where

y

1

=

x

1

/x

2

and

H(y) = 1

−

y

2

−

kx

−

α

−

2

2

(y

−

α

−

1).

Notethat

H

′

(y) = 0

onlyfor

y

=

y

∗

,where

(2.1)

y

∗

=

lx

−

1

2

,

with

l

= (αk/2)

1/(α+2)

and

H

′′

y

0

y

1

y

_∗

₁

H

Figure 2

Consequently,

H(y)

>

H

(y

∗

)

−

H

(y

1

)

y

∗

−

y

1

(y

−

y

1

) +

H

(y

1

),

y

∈

(y

1

, y

∗

),

H

(y)

>

H

(y

∗

)

−

H

(1)

y

∗

−

1

(y

−

1) +

H

(1),

y

∈

(y

∗

,

1),

(seeFigure 2)and

H

(y

1

) =

H

(1) = 0

imply

(2.2)

1

p

H

(y)

<

r

y

∗

−

y

1

H

(y

∗

)

1

√

_y

−

y

1

,

y

∈

(y

1

, y

∗

),

and

(2.3)

1

p

H

(y)

<

s

1

−

y

∗

H

(y

∗

)

1

√

1

−

y

,

y

∈

(y

∗

,

1).

It isalso

(2.4)

1

p

1

−

y

2

<

1

p

H(y)

,

y

∈

(y

1

,

1).

Due to

r

min

→

0

,

r

max

→ ∞

as

E

→ ∞

,itis

x

1

→

0+, x

2

→ ∞

,whene

(2.5)

y

1

→

0, y

∗

→

0

H

(y

∗

) = 1

−

y

2

∗

−

kl

−

α

_x

−

2

2

+

kx

−

α

−

2

2

→

1.

TheproofoftheLemmaisbaseduponthefollowinginequality

(2.6)

Z

1

y

1

dy

p

H

(y)

−

Z

1

y

1

dy

p

1

−

y

2

6

Z

1

y

∗

dy

p

H(y)

−

Z

1

y

∗

dy

p

1

−

y

2

+

Z

y

∗

y

1

dy

p

H(y)

+

Z

y

∗

y

1

dy

p

Evidently,

R

y

1

dy

√

1

−

y

2

→

π

2

and

R

∗

y

1

dy

√

1

−

y

2

→

0

. The inequalities (2.4) and (2.3),imply

Z

1

y

∗

dy

p

H

(y)

−

Z

1

y

∗

dy

p

1

−

y

2

=

Z

1

y

∗

p

1

−

y

2

₋

p

H

(y)

p

H(y)

p

1

−

y

2

dy

=

Z

1

y

∗

1

−

y

2

₋

_H

_(y)

p

H(y)

p

1

−

y

2

₍

p

H(y) +

p

1

−

y

2

₎

dy

=

kx

−

α

−

2

2

Z

1

y

∗

y

−

α

₋

₁

p

H

(y)

p

1

−

y

2

₍

p

H

(y) +

p

1

−

y

2

₎

dy

6

kx

−

α

−

2

2

Z

1

y

∗

y

−

α

₋

₁

H

(y)

3/2

dy

6

1

−

y

∗

H(y

∗

)

3/2

kx

−

α

−

2

2

Z

1

y

∗

y

−

α

₋

₁

1

−

y

1

√

1

−

y

dy.

Thelasttermtendsto

0

: indeed, owingto(2.5),itis

1

−

y

∗

H(y

∗

)

3/2

→

1,

and thanks to

y

−

α

₋

₁

_{< y}

−

α

₆

_y

−

α

∗

for

y

>

y

∗

, and the equality

x

2

y

∗

=

l

(see (2.1)),respetively,itfollows

kx

−

α

−

2

2

Z

1

y

∗

y

−

α

₋

₁

1

−

y

1

√

1

−

y

dy

=

kx

−

α

−

2

2

Z

1

1/2

y

−

α

₋

₁

1

−

y

1

√

1

−

y

dy

+

kx

−

α

−

2

2

Z

1/2

y

∗

y

−

α

₋

₁

1

−

y

1

√

1

−

y

dy

6

kx

−

α

−

2

2

Z

1

1/2

y

−

α

₋

₁

1

−

y

1

√

1

−

y

dy

+

kx

−

α

−

2

2

Z

1/2

y

∗

y

−

α

∗

(1

−

y

2

₎

3/2

dy

=

kx

−

α

−

2

2

Z

1

1/2

y

−

α

₋

₁

1

−

y

1

√

1

−

y

dy

+

kl

−

α

_x

−

2

2

Z

1/2

y

∗

dy

(1

−

y

2

₎

3/2

→

0.

Thus, the rst expression on the r.h.s. in (2.6) tends to

0

. It only remains to provethesamefor theseond termonr.h.s.of (2.6). Itis, however,asimple

onsequeneof (2.2):

Z

y

∗

y

1

dy

p

H(y)

6

r

_y

∗

−

y

1

H(y

∗

)

Z

y

∗

y

1

dy

√

_y

−

y

1

→

0.

Aknowledgments. I want to thank prof. Alain Albouy for his beneial

Referenes

1. A.Albouy,Leturesonthetwo-bodyproblem,ClassialandCelestialMehanis: TheReife

Letures,PrinetonUniversityPress,Prineton,2002,63116.

2. V.I.Arnold,Mathematialmethodsoflassialmehanis,Springer-Verlag,NewYork,1989.

3. ,Ordinarydierentialequations,MITPress,1978.

4. J.Bertrand, Théorème relatifau movementd'un point attiréversunentrexe,Comptes

rendus77(1873),849853.

5. G. Darboux, Étude d'une question relative au mouvement d'un point sur une surfae de

révolution,BulletindelaS.M.F.5(1877),100113.

6. J.Féjoz,L.Kazmarek,SurlethéorèmedeBertrand(d'aprèsMihaelHerman),Ergod.Th.

&Dynam.Sys.24(2004),17.

7. Y.Grandati,A.Bérard,F.Ménas,InverseproblemandBertrand'stheorem,Am.J.Phys.76

(2008),782787.

8. H.Liebmann,ÜberdieZentralbewegunginderniheuklidishenGeometrie,LeipzigBerihte

55(1903),146153.

9. I.Newton,Philosophiaenaturalisprinipiamathematia,London,1687.

10. V.Perlik,Bertrandspaetimes,Class.QuantumGrav.9(1992),10091021.

11.

O. A. Zagrdski, E. A. Kudrvceva, D. A. Fedoseev

,

Obobwenie teorem y Bertrana

O DOKAZU BERTRANOVE TEOREME

Rezime.

_{U ovom radu dopunjujemo dokaze Bertranove teoreme iz}

Bertrano-vog originalnog rada

Théorème relatif au movement d'unpoint attiré vers un

entre xe

, i Arnoljdove knjige Matematiqke metode klasiqne mehanike,

predstavljajui nedostajue korake koji omoguuju da se elastiqni i

gravitacioni potencijal izdvoje meu potencijalima oblika stepene

fu-nkcije.

FaultyofSienesandMathematis (Reeived27.04.2015)

UniversityofBanjaLuka (Revised26.06.2015)

BanjaLuka

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